JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

Select Answer:

Visualized Solution

Understanding the Group Structure

Total Boys = $5$
Total Girls = $n$
Team Size = $3$
Constraint: At least one boy and at least one girl.

Analyzing the Constraints

Team size is $3$ .
Must include $\geq 1$ Boy and $\geq 1$ Girl.
This leads to two distinct mutually exclusive cases.

Case 1: One Boy and Two Girls

Case 1: Select $1$ Boy and $2$ Girls.
Total members = $1 + 2 = 3$ .
Satisfies the "at least one" constraint.

Case 2: Two Boys and One Girl

Case 2: Select $2$ Boys and $1$ Girl.
Total members = $2 + 1 = 3$ .
Also satisfies the constraint.

Setting up the Combination Formula

Ways for Case 1: $^{5} C_{1} \cdot^{n} C_{2}$
Ways for Case 2: $^{5} C_{2} \cdot^{n} C_{1}$
Total Ways = $^{5} C_{1} \cdot^{n} C_{2} +^{5} C_{2} \cdot^{n} C_{1} = 1750$

Expanding the Combinations

$^{5} C_{1} = 5$
$^{n} C_{2} = \frac{n ( n - 1 )}{2}$
$^{5} C_{2} = \frac{5 \cdot 4}{2} = 10$
$^{n} C_{1} = n$

Substituting the Expanded Values

Equation: $5 \cdot \frac{n ( n - 1 )}{2} + 10 \cdot n = 1750$

Simplifying the Equation

Multiply the entire equation by $2$ to remove the fraction.
$5 n (n - 1) + 20 n = 3500$

Expanding the Brackets

Expand $5 n (n - 1)$ : $5 n^{2} - 5 n$
Substitute back: $5 n^{2} - 5 n + 20 n = 3500$

Forming the Quadratic Equation

Combine like terms: $- 5 n + 20 n = 15 n$
Equation: $5 n^{2} + 15 n = 3500$
Divide by $5$ : $n^{2} + 3 n - 700 = 0$

Factoring the Quadratic

We need two numbers that multiply to $- 700$ and add to $3$ .
These numbers are $28$ and $- 25$ .
Split the middle term: $n^{2} + 28 n - 25 n - 700 = 0$
Factorize: $(n - 25) (n + 28) = 0$

Solving for n

Possible values: $n = 25$ or $n = - 28$
Since $n$ represents the number of girls, $n$ must be a positive integer.
Therefore, $n = 25$ .

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

My dear student, welcome to a fascinating exploration of combinatorics. Often, students fear these problems, seeing them as a maze of formulas. But I want you to see them as a story of possibilities.

Imagine you are standing in front of a group of students: 5 boys in one corner and

n

girls in another. You are tasked with forming a team of 3.

There is a catch—a constraint that forces us to think. We need at least one boy and at least one girl. This isn't just a math problem; it is a logic puzzle.

The Logic of Cases

Since our team size is fixed at 3, how can we satisfy the condition of having at least one of each gender? We cannot have all boys, and we cannot have all girls.

This leaves us with only two mutually exclusive scenarios. Either we pick 1 boy and 2 girls, or we pick 2 boys and 1 girl. These are our two paths.

In the first path, we select 1 boy from 5 and 2 girls from

n

. Mathematically, this is

^{5} C_{1} \cdot^{n} C_{2}

In the second path, we select 2 boys from 5 and 1 girl from

n

, which is

^{5} C_{2} \cdot^{n} C_{1}

. Because these paths are distinct, we add them together to find the total number of ways:

^{5} C_{1} \cdot^{n} C_{2} +^{5} C_{2} \cdot^{n} C_{1} = 1750

Peeling Back the Algebra

Now, let's bring in the machinery of combinations. We know that

^{5} C_{1} = 5

and

^{5} C_{2} = 10

. For the girls,

^{n} C_{2} = \frac{n ( n - 1 )}{2}

and

^{n} C_{1} = n

Substituting these into our equation, we get:

5 \cdot \frac{n ( n - 1 )}{2} + 10 n = 1750

To make this equation elegant, let's multiply everything by 2 to clear the denominator:

5 n (n - 1) + 20 n = 3500

Now, expand the bracket:

5 n^{2} - 5 n + 20 n = 3500

Simplifying this, we arrive at

5 n^{2} + 15 n = 3500

. Dividing the entire equation by 5 gives us the quadratic:

n^{2} + 3 n - 700 = 0

The Final Resolution

We are looking for two numbers that multiply to

- 700

and add to

3

. Through a bit of mental arithmetic, we find

28

and

- 25

Thus, the equation factors beautifully into:

(n - 25) (n + 28) = 0

This gives us two potential values for

n

25

and

- 28

. As we discussed, we must reject the negative value because we cannot have a negative number of students.

Therefore, $n = 25$ . You have successfully navigated the constraints, translated the logic into algebra, and solved for the unknown.

Common Traps & FAQs

Wait, why can't we just use the total number of ways to pick 3 students and subtract the cases with no boys or no girls?

That is actually a brilliant alternative approach! You could calculate the total ways to pick 3 students from

(5 + n)

and subtract the cases where you pick 3 boys (which is

^{5} C_{3}

) and the cases where you pick 3 girls (which is

^{n} C_{3}

). However, direct case analysis is often less prone to calculation errors in the exam hall. Both methods will lead you to the same beautiful result of

n = 25

Why do we multiply the combinations within a case, like

^{5} C_{1} \cdot^{n} C_{2}

Great question! We multiply because these are independent, sequential choices that form a single, unified team. For every single way you pick a boy, you have a set number of ways to pick the girls. This is the Fundamental Principle of Counting in action. Think of it as a tree diagram: for each branch of 'boy selection', there are multiple branches of 'girl selection'.

Why do we reject the negative root in the quadratic equation?

Mathematics is the language of the universe, but it doesn't always know the physical context of your problem. In this case,

n

represents the number of girls in a group. A count of people must be a positive integer. While

- 28

is a valid algebraic solution to the quadratic equation

n^{2} + 3 n - 700 = 0

, it is physically impossible to have negative twenty-eight girls. Always check your roots against the physical constraints of the problem!

Select Answer:

Visualized Solution

Understanding the Group Structure

Analyzing the Constraints

Case 1: One Boy and Two Girls

Case 2: Two Boys and One Girl

Setting up the Combination Formula

Expanding the Combinations

Substituting the Expanded Values

Simplifying the Equation

Expanding the Brackets

Forming the Quadratic Equation

Factoring the Quadratic

Solving for n

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Logic of Cases

Peeling Back the Algebra

The Final Resolution

Similar Questions

There are ten boys $B_{1}, B_{2}, \dots, B_{10}$ and five girls $G_{1}, G_{2}, \dots, G_{5}$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_{1}$ and $B_{2}$ together should not be the members of a group, is ______.

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

A class contains $b$ boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $b + 3 g$ is equal to ______.

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

List-I

List-II

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

Select Answer:

Visualized Solution

Understanding the Group Structure

Analyzing the Constraints

Case 1: One Boy and Two Girls

Case 2: Two Boys and One Girl

Setting up the Combination Formula

Expanding the Combinations

Substituting the Expanded Values

Simplifying the Equation

Expanding the Brackets

Forming the Quadratic Equation

Factoring the Quadratic

Solving for n

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Logic of Cases

Peeling Back the Algebra

The Final Resolution

Similar Questions

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b+3g is equal to ______.

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

List-I

List-II

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

A class contains $b$ boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $b + 3 g$ is equal to ______.